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Control Systems

Laplace domain analysisL. Lanari

Monday 26 October 15

Lanari: CS - Laplace domain analysis 2

outline

• introduce the Laplace unilateral transform

• define its properties

• show its advantages in turning ODEs to algebraic equations

• define an Input/Output representation of the system through

the transfer function

• exploring the structure of the transfer function

• solve a realization problem



Laplace transform

f(t) F (s)

t 2 R s 2 C

complex valued function in the complex variable s

F (s) =

Z 1

0f(t)e�stdt Laplace (unilateral)

transform

Region of existence: for all s with real part greater equal to an abscissa of convergence ¾0

for f(t) with no impulse and no discontinuities at t = 0

L

F (s) = L[f(t)]

real valued function in the real variable t

Re

Im

f(t) = e�at, a > 0, �0 = �a

-a region of convergence

example



inverse Laplace transform

f(t) = L�1[F (s)] =1

2�j

Z ��j1

��j1F (s)estds

f(t) F (s)L

L�1

unique for one-sided functions f(t) = 0 for t < 0f(t) t � 0defined for or

(one-to-one)with this assumption the inverse

Laplace transform is unique

0

t

f(t)

F(s)

all give the same F(s)

?



Laplace transform

F (s) =

Z 1

0�f(t)e�stdt

L[�(t)] =Z 1

0��(t)e�stdt =

Z 1

�1�(t)e�stdt = e�s0 = 1

Laplace transform of the impulse

Linearity L[�f(t) + ⇥g(t)] = �L[f(t)] + ⇥L[g(t)]

correct definition of the Laplace transform

Ldf(t)

dt

�= sL[f(t)]� f(0)Derivative property

can be applied iteratively Lhf(t)

i= s2L[f(t)]� sf(0)� f(0) very useful in

model derivation



LTI systems

also useful here x(t) = Ax(t) +Bu(t)

X(s) = (sI �A)�1x0 + (sI �A)�1

B U(s)

algebraic solution

and therefore

Y (s) = C(sI �A)�1x0 +

⇥C(sI �A)�1

B +D

⇤U(s)

(proof): linearity + derivative property

x(0) = x0



LTI systems

X(s) = (sI �A)�1x0 + (sI �A)�1

B U(s)

and therefore, comparing

x(t) = eAtx0 +

Z t

0eA(t��)Bu(�)d�

L⇥eAt

⇤= (sI �A)�1

L⇥eat

⇤=

1

s� aL [��1(t)] =

1

s

Heaviside step function

��1(t)

1

t

1 for t � 0

0 for t < 0

state ZIRtransform

state ZSRtransform

| {z } | {z }



LTI systemsy(t) = CeAtx0 +

Z t

0w(t� �)u(�)d�

w(t) = CeAtB +D �(t)

W (s) = C(sI �A)�1B +D

Y (s) = C(sI �A)�1x0 +

⇥C(sI �A)�1

B +D

⇤U(s)

= C(sI �A)�1x0 +W (s) U(s)

with

transform

LZ t

0w(t� �)u(�)d�

�= W (s) U(s)

Convolution theorem

being L[�(t)] = 1 the output response transform corresponding to u(t) = ±(t) is indeed W(s)

same for H(s)



transfer function

Input/Output behavior

x0 = 0 (ZSR)

yZS(t) =

Z t

0w(t� �)u(�)d�

YZS(s) = W (s) U(s)

W (s) = C(sI �A)�1B +D

=YZS(s)

U(s)

= L [w(t)]

Input/Output behavior independent from state choice?

independent from state space representation?

Transfer function



(A,B,C,D)

z = Tx det(T ) 6= 0

( eA, eB, eC, eD)

z = eAz + eBu

y = eCz + eDu

eA = T AT�1 eB = T B eC = C T�1 eD = D

transfer function

x = Ax+Bu

y = Cx+Du

W (s) = C(sI �A)�1B +D = eC(sI � eA)�1 eB + eD

for a given system, the transfer function is unique



shape of the transfer function

W (s) = C(sI �A)�1B +D

inverse through the adjoint (transpose of the cofactor matrix)

1

det(sI �A)

C (adjoint of (sI �A))B +D

cofactor(i, j) = (�1)

i+jminor(i, j)

1

det(sI �A)

C (cofactor of (sI �A))

T B +D

polynomial of order n - 1

polynomial of order n

polynomial of order n - 1

rational function (strictly proper)

strictly proper rational function: degree of numerator < degree of denominatorproper rational function: degree of numerator = degree of denominator



shape of the transfer function

proper rational function

W (s) = strictly proper rational function + D

W (s) =N(s)

D(s)

D = 0

D 6= 0 proper rational function

strictly proper rational function

W (s) =N(s)

D(s) poleszeros

(for coprime N(s) & D(s))roots

i.e. no common roots

from previous analysis the poles are a subset of the eigenvalues of A

{poles} {eigenvalues}more on this later

✓



partial fraction expansion

Let be a strictly proper rational function with

coprime N(s) and D(s) and distinct roots of D(s)F (s) =

N(s)

D(s)

D(s) = an

nY

i=1

(s� pi) ) F (s) =N(s)

anQn

i=1(s� pi)

F (s) =nX

i=1

Ri

s� pi

Ri = [(s� pi)F (s)]��s=pi

i.e.

then F(s) can be expanded as

with the residues Ri computed as

(distinct roots case)

this result will be used for the• transfer function W(s)

• output zero-state response Y(s)



W (s) =N(s)

D(s)

poles & eigenvalues

W (s) =1

det(sI �A)N 0(s)

from def?

if det(sI �A) and N 0(s) coprime {poles} = {eigenvalues}

characteristicpolynomial

if det(sI �A) and N 0(s) not coprime {poles} subset of {eigenvalues}

Input/Outputrepresentation

Input/State/Outputrepresentation

vs

here visible

Transferfunction

Statespace

(

we need to understand when & why this happens(so to understand when we can consider the transfer function equivalent to a state space representation)



we have a “hidden mode” associated to the eigenvalue ¸j(see structural properties)

poles & eigenvalues (distinct eigenvalues of A case)

n = state space dimension = dimension of A = number of eigenvaluesnp = number of poles in W(s)

W (s) =N(s)

D(s)=

N(s)Qnp

i=1(s� pi)=

npX

i=1

Ri

s� pi

if the eigenvalue ¸j does not appear as a poleand/or

vTj B = 0

Cuj = 0

W (s) = L[w(t)] = L[CeAtB] = L[C

0

@nX

j=1

e�jtujvTj

1

AB] =nX

j=1

Cuj vTj B

s� �j

done the same analysis in t for n = 2

NB distinct eigenvalues is different from diagonalizable

partial fraction expansion

spectral form




eAtB = H(t)

CeAt = �(t)

vTj B = 0

Cuj = 0

implies the corresponding mode will not appear in the state impulsive response

the corresponding mode (or eigenvalue) is said to be uncontrollable

implies the corresponding mode will not appear in the output transition matrix

the corresponding mode (or eigenvalue) is said to be unobservable

If for an eigenvalue ¸j we have that

all the natural modes appear in the state ZIR for some generic initial condition eAtfact:




TheoremEvery pole is an eigenvalue.

An eigenvalue ¸i becomes a pole if and only if it is both controllable and observable

or equivalently the following two PBH rank tests are both verified

vTi B 6= 0 Cui 6= 0

rank�A� �iI B

�= n rank

A� �iI

C

!= n

Popov-Belevitch-Hautuscontrollability test

Popov-Belevitch-Hautusobservability test

and

(the PBH test could be tested for a generic ¸ but matrix A - ¸I loses rank only for ¸=¸i )

and




Where does the PBH test comes from?

Observability (sketch):

rank

A� �iI

C

!< n

means that the rectangular (n+1) x n matrix has not full column rank and therefore it has a non-zero nullspace, that is there exists a n vector ui such that

A� �iI

C

!ui = 0

(A� �iI)ui = 0 Aui = �iui

Cui = 0 Cui = 0

( (

that is there exists an eigenvector which belongs to the nullspace of C (or the corresponding mode is unobservable)



example

A =

✓�1 10 1

◆B =

✓10

◆C =

�0 1

�

(sI �A)�1 =

✓ 1s+1

1(s+1)(s�1)

0 1s�1

◆= M1

1

s+ 1+M2

1

s� 1

M1 =⇥(s+ 1)(sI �A)�1

⇤s=�1

=

✓1 �1/20 0

◆

M2 =⇥(s� 1)(sI �A)�1

⇤s=1

=

✓0 1/20 1

◆

fractional decompositionworks also for rational matrices

with

eAt = M1e�t +M2e

ta different way to compute the matrix exponential

both natural modes appear (as it should be) in the state transition matrix



(sI �A)�1B =

✓1

s+10

◆

C(sI �A)�1 =�0 1

s�1

�

u1 / (A� �1I)u1 = 0

0 10 2

�u1 = 0 u1 =

10

�

u2 / (A� �2I)u2 = 0

�2 10 0

�u2 = 0 u2 =

12

�vT1 =

⇥1 �1/2

⇤

vT2 =⇥0 1/2

⇤

Cu1 = 0 Cu2 6= 0vT1 B 6= 0 vT2 B = 0

mode corresponding to ¸2 does not appear

mode corresponding to ¸1 does not appear

equivalently

eAtB

w(t) = CeAtB = 0

CeAt

forced response will always be zero independently from the input applied

(look at the 2 first order ODE)

example

W (s) = 0

no poles



example equivalently with the PBH rank test

rank

✓0 1 10 2 0

◆= 2 = n

rank

✓�2 1 10 0 0

◆= 1 < n

rank

0

B@0 10 2

0 1

1

CA = 1 < n rank

0

B@�2 10 0

0 1

1

CA = 2 = n

controllability test

observability test

mode corresponding to ¸2 is uncontrollable

mode corresponding to ¸1 is controllable

mode corresponding to ¸1

is unobservablemode corresponding to ¸2

is observable



poles & eigenvalues (general case)

TheoremEvery pole is an eigenvalue.

An eigenvalue ¸i becomes a pole with the multiplicity ma (algebraic multiplicity) if and only if

both PBH rank tests are verified

rank�A� �iI B

�= n rank

A� �iI

C

!= n

NB - If one of the two conditions is not verified then the eigenvalue ¸i will appear as a pole with multiplicity strictly less than the algebraic multiplicity, possibly even 0 (in this case we will have a hidden eigenvalue). In particular the eigenvalue will appear at most as a pole with multiplicity equal to its index (dimension of the largest Jordan block).

controllability observability

NB - If for an eigenvalue ¸i the geometric mg(¸i) >1 then there exists a hidden dynamics.



example

A0 =

2

4�1 1 00 �1 10 0 �1

3

5

(sI �A0)�1 =

1

(s� �1)3

2

64(s� �1)2 (s� �1) 1

0 (s� �1)2 (s� �1)

0 0 (s� �1)2

3

75

B =

2

4886= 0

3

5 C =⇥6= 0 8 8

⇤

B1 =

2

4001

3

5 C1 =⇥1 0 0

⇤F1(s) =

1

(s� �1)3

B2 =

2

4100

3

5 C2 = C1 =⇥1 0 0

⇤F2(s) =

(s� �1)2

(s� �1)3=

1

s� �1

B3 = B2 =

2

4100

3

5 C3 =⇥0 0 1

⇤F3(s) = 0

PBH rank testverified for

easily seen from A0 � �1I =

2

40 1 00 0 10 0 0

3

5

index of ¸1



A4 =

2

4�1 1 00 �1 00 0 �1

3

5 A4 � �1I =

2

40 1 00 0 00 0 0

3

5

(sI�A4)�1 =

1

(s� �1)3

2

4(s� �1)2 (s� �1) 0

0 (s� �1)2 00 0 (s� �1)2

3

5 =1

(s� �1)2

2

4(s� �1) 1 0

0 (s� �1) 00 0 (s� �1)

3

5

B4 =

2

4010

3

5 C4 =⇥1 0 0

⇤F4(s) =

1

(s� �1)2

B5 =

2

4100

3

5 C5 = C1 =⇥1 0 0

⇤

B6 = B1 =

2

4001

3

5 C6 = C1 =⇥1 0 0

⇤F6(s) = 0

F5(s) =s� �1

(s� �1)2=

1

s� �1

example

since

the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of ¸1 = 2



example

A7 =

2

4�1 0 00 �1 00 0 �1

3

5 (sI �A7)�1 =

1

s� �1

2

41 0 00 1 00 0 1

3

5 etc ...

the PBH rank test will never be satisfied independently from B and C. At most the eigenvalue will appear as a pole with multiplicity = index of ¸1 = 1



Laplace transform tables

�(t)

��1(t)1

s

tk

k!

1

sk+1

eat1

s� a

sin�t�

s2 + �2=

1/2j

s� j�� 1/2j

s+ j�

cos�ts

s2 + �2=

1/2

s� j�+

1/2

s+ j�

sin(�t+ ⇥)s sin⇥+ � cos⇥

s2 + �2

1

tk

k!eat

1

(s� a)k+1

eat sin�t

eat cos�t

�

(s� a)2 + �2

(s� a)

(s� a)2 + �2



(A,B,C,D) W (s) = C(sI �A)�1B +D

how ?

ok

realizations

(A,B,C,D)W (s) = C(sI �A)�1B +D

infinite solutions

• state dimension ?

• how can we easily find one state space representation (A, B, C, D) ?

• may be complicated for MIMO systems (here SISO)

• we see only one, obtainable directly from the coefficients of the transfer

function (others are obtainable by simple similarity transformations)

realization



realizations

W (s) =N(s)

D(s)

W (s) =bn�1sn�1 + bn�2sn�2 + · · ·+ b1s+ b0

sn + an�1sn�1 + an�2sn�2 + · · ·+ a1s+ a0+D

coprime N(s) & D(s)

- find D

- one possible realization for A, B, C

Ac =

2

66664

0 1 0 · · · 00 0 1 · · · 0· · · · · · · · · · · · · · ·0 0 0 · · · 1

�a0 �a1 �a2 · · · �an�1

3

77775Bc =

2

666664

00...01

3

777775

Cc =⇥b0 b1 b2 · · · bn�1

⇤

- state with dimension n

controller canonical form(useful for eigenvalue assignment)


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